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I have $2$ following problems. Find integer roots of

$$\begin{align} &1)~\frac{x+y}{x^2-xy+y^2}=\frac3z \\ &2)~x^3y^3-4xy^3+y^2+x^2-2y-3=0 \end{align}$$

I have no idea to solve them. I try to guess roots of the second, they are $\left( -2, 1\right), \left( 0, -1\right), \left( 0, 3\right), \left( 2, 1\right) $. Please help me. Thank you very much.

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  • $\begingroup$ These problems seem very interesting. Where did you get them from? $\endgroup$ – Toby Mak Dec 8 '18 at 9:39
  • $\begingroup$ @TobyMak These are my homework. $\endgroup$ – Cglkttca Dec 8 '18 at 9:42
  • $\begingroup$ What class are you taking? $\endgroup$ – Toby Mak Dec 8 '18 at 9:42
  • $\begingroup$ @TobyMak I'm in grade 9. $\endgroup$ – Cglkttca Dec 8 '18 at 9:43
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    $\begingroup$ What lesson is this that you were given that exercise ? $\endgroup$ – Rebellos Dec 8 '18 at 10:03
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Question 1:

\begin{align} & \frac{x+y}{x^2-xy+y^2} = \frac 3z \\ \iff & \frac{3(x^2-xy+y^2)}{x+y} = z\\ \iff & 3(x+y) - \frac{9xy}{x+y} = z \\ \iff & \frac{9xy}{x+y} = 3x+3y-z \\ \implies & \frac{9}{\frac 1x + \frac 1y} = 3x+3y-z \end{align}

From the LHS we see that $\frac 1x + \frac 1y$ must be equal to $\pm 1, \pm 3, \pm 9$ since the RHS is an integer. But of course, since $x,y$ are integers, $\frac 1x + \frac 1y \in [-2,2]$ and it follows that $\frac 1x + \frac 1y = \pm 1$.

Moreover, the only way this can happen is if $x = y = \pm 2$.

Hence, the only solutions are $(x,y,z) = (2,2,3)$ and $(x,y,z) = (-2,-2,-3)$.

EDIT:

As pointed out in the comments, I have made the mistake of dividing by $0$, so I have changed the last $\iff$ to an $\implies$.

If $x=0$, then $z=3y$ so we get the solutions $(x,y,z) = (0,t,3t)$

Similarly when $y=0$ we get the solutions $(x,y,z) = (t,0,3t)$ for any $t \in \Bbb Z$ with $t \neq 0$.

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  • $\begingroup$ I got it. Thank you very much. $\endgroup$ – Cglkttca Dec 8 '18 at 10:20
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    $\begingroup$ It's a nice answer, however, you forgot the cases when x and y are 0. $\endgroup$ – Ankit Kumar Dec 8 '18 at 10:29
  • $\begingroup$ Yes, in this case, we get more solutions $\endgroup$ – Cglkttca Dec 8 '18 at 10:34

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